What mass of ethylene glycol C2H6O2, the main component of antifreeze, must be added to 10.0 L water to produce a solution for use in a car's radiator that freezes at -23C? Assume density for water is exactly 1g/mL.
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This is what i have so far:
Freezing point for water is 0.0degree C and Kf = 1.86C/m.
First we find out the freezing point:
=(-23.0)-(0.0_
=-23.0 degrees C
Calculate target Colligative Molarity:
= -23=1.86 * Cm
= 12.37= Cm.
Now we determine the number of grams needed to produce a solution with Cm= 12.37.
We need to find i* molarity:
I don't know how to determine the i: i know that its an integer equal to the number of particles(ions) present in one formula unit of the solute. If i is greater than 1 the molecules dissociates; and if less than 1 then the molecule associates in solution.
and then i use:
molarity = moles of solute (KNO3)/ kg of solvent (water)
SO:
First we find out the freezing point:
=(-23.0)-(0.0_
=-23.0 degrees C
Calculate target Colligative Molarity:
= -23=1.86 * Cm
= 12.37= Cm.
i* molality
i= 1
12.37= 1*m
12.37= m
molality= moles of solute(C2H6O2)/kg of solvent water
12.37m= moles C2H6O2/ 10.0kg water
123.7= moles of C2H6O2
123.7moles C2H6O2 * 62.26g C2H6O2
= 7701.56 g of C2H6O2
I think this number is quite high can someone please double check to see if i got the answer right.
Thank YOu
CORRECTION : molarity = moles of solute (C2H6O2)/ kg of solvent (water)
i is 1 in antifreeze.
You are not using molarity, you should be using molality, which is moles of antifreeze/kg of solvent
I don't know how potassium nitrate got in there.
12.37=molesantifreeze/kg water
solve for the moles of antifreeze, then convert that to mass
Before I forget, molarity is the number of moles per liter of solution. Molality is the number of moles per kg of solvent. I think your last statement is incorrect. The i for ethylene glycol is 1; i.e., it doesn't ionize.
thats rite
thats not right its n
Naomi is incorrect, assuming density is 1g/mL, that's equal to 1kg/L. So 10.0L water = 10.0kg water.
Saira's answer is correct.
To determine the value of i, you need to know the chemical formula of the solute, in this case, ethylene glycol (C2H6O2). Ethylene glycol does not dissociate or ionize in water, so it does not produce ions. Therefore, its i value is 1.
Now, let's calculate the molarity of ethylene glycol needed to achieve a freezing point of -23°C. We'll use the formula:
Cm = i * molarity
Substituting the given values, we have:
12.37 = 1 * molarity
So, the molarity of ethylene glycol needed is 12.37 mol/L.
To determine the mass of ethylene glycol needed, we'll use the equation:
moles of solute = molarity * volume of solvent
The volume of solvent is given as 10.0 L. Converting this to liters:
10.0 L * (1000 mL/1 L) = 10,000 mL
Now, let's calculate the moles of ethylene glycol:
moles of solute = 12.37 mol/L * 10,000 mL
Note that we convert mL to liters to match the units of molarity.
moles of solute = 0.1237 mol
Finally, to determine the mass of ethylene glycol, we'll use the molar mass of C2H6O2:
Molar mass of C2H6O2 = (2 * atomic mass of C) + (6 * atomic mass of H) + (2 * atomic mass of O)
= (2 * 12.01 g/mol) + (6 * 1.008 g/mol) + (2 * 16.00 g/mol)
= 62.07 g/mol
Now, let's calculate the mass:
mass = moles of solute * molar mass of C2H6O2
mass = 0.1237 mol * 62.07 g/mol
mass ≈ 7.67 g
Therefore, approximately 7.67 grams of ethylene glycol must be added to 10.0 L of water to produce a solution with a freezing point of -23°C.